Total output for each 24 hour period is 7 units The utility produces

Total output for each 24 hour period is 7 units the

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during the night only 3 units are demanded. Total output for each 24-hour period is 7 units. The utility produces electricity according to the production function y i = ( KF i ) 1 / 2 , i = day, night , where K is the size of the generating plant, and F i is tons of fuel. The °rm must build a single plant; it cannot change plant size from day to night. If plant size costs w k per 24-hour period and a ton of fuel costs w f , what size plant will the utility build? Fuel needed per day and per night: F d = y 2 d K F n = y 2 n K Total costs per day (24 hours) are: TC = w k K + w f ( F d + F n ) min K ! Choose plant size to minimize cost: TC = w k K + w f y 2 d + y 2 n K min K ! FOC: w k + w f y 2 d + y 2 n K 2 ( - 1) ! = 0 w f ( y 2 d + y 2 n ) d 1 K 2 = w k K = s w f ( y 2 d + y 2 n ) w k Since demand is y d = 4 and y n = 3 , K = r w f (16 + 9) w k = r 25 w f w k = 5 r w f w k . 13
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8. Suppose that a °rm has production function Q = 2 L K . Remark: Cobb-Douglas production function with A = 2 , α L = 1 , α K = 1 2 . a) If the price of labour is 2 and the price of capital is 4, what is the optimal input mix? Now we have p L = 2 , p K = 4 . Cost minimisation program: min K,L 2 · L + 4 · K s.t. (2 L K - Q ) = 0 ⇒ L = 2 L + 4 K - λ (2 L K - Q ) FOC: L ∂K = 4 - 1 2 · 2 · L K λ ! = 0 λ = 4 K L L ∂L = 2 - 2 λ K ! = 0 λ = 1 K 4 K L = 1 K 4 K = L This last result describes the optimal input mix. Insert this into the constraint: 2 L K - Q = 0 8 K K - Q = 0 8 K 3 / 2 - Q = 0 K = 3 s Q 8 2 = Q 8 2 / 3 L = 4 · Q 8 2 / 3 Summary: ° slope of isocost line: p L p K = 1 2 ° slope of isoquant: - ∂Q/∂L ∂Q/∂K = 2 K L K = 2 K L In the optimum, the isoquant touches the isocost line, i. e. their slopes are equal: 1 2 = 2 K L L = 4 K This is the same condition as above. b) For an output level of Q = 1000 , how much of each input will be used? see a): K = Q 8 2 / 3 = 1 , 000 8 2 ! 1 / 3 = 3 r 1 , 000 , 000 64 = 100 4 = 25 L = 4 · K = 100 14
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c) How should inputs be scaled to attain Q = 8000 ? What will be the percentage increase in cost? f ( sK, sL ) = 2 · ( s · L ) sK ! = 8 , 000 s 3 / 2 (2 L K ) | {z } for Q =1 , 000 = 8 , 000 s 3 / 2 = 8 s = 8 2 / 3 = 3 8 2 = 4 where we know from part (b) that 2 L K is the input mix that yields output Q = 1 , 000 at minimum cost. With this production function, quadruple labour and capital use to obtain 8,000 units of out- put. Due to s 3 / 2 we have increasing returns to scale. They are increasing because α L = 1 and α K = 1 2 , so α L + α K = 3 2 > 1 . 15
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